📚 Arithmetic Sequences: 1, 4, 7, … | 等差数列：1, 4, 7, …
In IGCSE Mathematics, understanding sequences and series is essential for building algebraic thinking. The pattern 1, 4, 7, 10, 13, ... is one of the most common examples of an arithmetic sequence (also called an arithmetic progression). This article will guide you through every key concept related to this simple yet powerful number pattern, from finding the nth term to calculating the sum of the first n terms, with clear bilingual explanations and step-by-step methods.
在 IGCSE 数学中，理解数列与级数是培养代数思维的基础。1, 4, 7, 10, 13, ... 这个模式是等差数列（也称算术数列）最常见的例子之一。本文将带你逐一掌握与这个简单而强大的数字模式相关的每一个核心概念，从求第n项到计算前n项和，提供清晰的双语说明和逐步解法。
1. Recognising the Pattern | 模式识别
Look at the list of numbers: 1, 4, 7, 10, 13, ... Each term is 3 more than the previous term. This constant difference is the defining feature of an arithmetic sequence. You can describe the rule as 'start at 1 and add 3 each time'.
观察这列数字：1, 4, 7, 10, 13, ... 每一项都比前一项大3。这种恒定的差值正是等差数列的标志性特征。我们可以将规律描述为"从1开始，每次加3"。
In general, a sequence is arithmetic if the difference between consecutive terms is always the same. For this sequence, the difference is 3. Not all sequences have a constant difference; for example, 2, 4, 8, 16, ... does not, because the gaps are 2, 4, 8, which are not constant.
一般来说，若相邻两项的差始终相同，则该数列就是等差数列。对于这个数列，差值是3。并非所有数列都有固定的差；例如2, 4, 8, 16, ... 就没有，因为差值分别为2, 4, 8，并不恒定。
2. First Term and Common Difference | 首项与公差
oted by d. In our pattern 1, 4, 7, ..., we have a = 1 and d = 3. These two values form the 'DNA' of the sequence.
对于任何一个等差数列，两个数值便足以完整定义它：首项，通常记作 a，和公差，记作 d。在模式 1, 4, 7, ... 中，a = 1，d = 3。这两个值构成了数列的"DNA"。
, 7 − 4 = 3, and 10 − 7 = 3. The common difference can be positive, negative or even zero.
检查公差时，应用后一项减去前一项：d = 第二项 − 第一项 = 4 − 1 = 3。同样，7 − 4 = 3，10 − 7 = 3。公差可以是正数、负数，甚至为零。
| Term (n) | Value | Difference |
|---|---|---|
| 1 | 1 | — |
| 2 | 4 | 4 − 1 = 3 |
| 3 | 7 | 7 − 4 = 3 |
| 4 | 10 | 10 − 7 = 3 |
3. The nth Term Formula | 第n项公式
− 1)d. The subscript n simply indicates the position in the sequence. You multiply the common difference by (n − 1), not by n, because the first term already has zero differences added.
首项为 a、公差为 d 的等差数列的第n项公式为：aₙ = a + (n − 1)d。下标 n 表示该数在数列中的位置。需要将公差乘以 (n − 1) 而不是 n，因为第一项已经加了零个公差。
Substituting a = 1 and d = 3 gives a powerful simplified form:
代入 a = 1 和 d = 3，可得一个简洁实用的形式：
aₙ = 1 + (n − 1) × 3 = 1 + 3n − 3 = 3n − 2
Thus, the nth term of the sequence 1, 4, 7, ... is 3n − 2 ng the entire list. For instance, the 5th term is 3×5 − 2 = 13, which matches the pattern.
因此，序列 1, 4, 7, ... 的第n项为 3n − 2。这个线性表达式使得你可以直接求出任意一项，而无需生成整个列表。例如，第5项为 3×5 − 2 = 13，与模式吻合。
4. Finding a Specific Term | 查找特定项
Once you have the nth term formula, finding any term is just a matter of substituting the position number n. For example, to find the 20th term, replace n by 20:
一旦得到第n项公式，求任意一项只需代入位置序号 n 即可。例如，求第20项，将 n 替换为20：
a₂₀ = 3 × 20 − 2 = 60 − 2 = 58
×100 − 2 = 298. The formula lets you jump deep into the sequence without tedious counting.
第100项则为 a₁₀₀ = 3×100 − 2 = 298。这个公式让你无需繁琐地逐个数数，就能直达序列深处。
This approach highlights the power of algebra in mathematics: a single equation replaces an entire table of values. The expression 3n − 2 is an arithmetic sequence generator.
这种方法凸显了代数在数学中的力量：一个简单的方程取代了整张数值表。表达式 3n − 2 就是一个等差数列生成器。
5. Is a Number in the Sequence? | 判断某个数是否在数列中
To determine whether a given number belongs to the sequence, set the nth term formula equal to that number and solve for n. If n is a positive integer, the number is a term of the sequence. For example, is 100 in the sequence?
要判断某个给定的数是否属于该数列，令第n项公式等于该数并求解 n。若 n 是正整数，则该数就是序列中的一项。例如，100 是否在该数列中？
3n − 2 = 100 → 3n = 102 → n = 34
n − 2 = 101 → 3n = 103 → n = 103/3 ≈ 34.33, not an integer, so 101 is not in the sequence.
因为 n = 34 是整数，所以100是该数列的第34项。再试101：3n − 2 = 101 → 3n = 103 → n = 103/3 ≈ 34.33，不是整数，因此101不在数列中。
ive n do not represent valid positions in the sequence 1, 4, 7, ... (which starts at n=1).
这类检验经常出现在考题中。关键是看 n 是否为正整数；n = 0 或负值不代表数列 1, 4, 7, ... 中的有效位置（该数列从 n=1 开始）。
6. Sum of an Arithmetic Series | 等差数列求和
When the terms of a sequence are added together, we get a series. The sum of the first n terms of an arithmetic sequence is given by two equivalent formulas:
将数列各项相加，便得到一个级数。等差数列前 n 项和的公式有以下两种等价形式：
Sₙ = n/2 × (2a + (n − 1)d) or Sₙ = n/2 × (a + l)
first 10 terms. First, find the 10th term: l = 3×10 − 2 = 28. Then use the a + l version:
其中 l 为末项（即第n项）。对于 1, 4, 7, ...，a = 1, d = 3，我们可以求前10项的和。首先求出第10项：l = 3×10 − 2 = 28。然后使用 a + l 版公式：
S₁₀ = 10/2 × (1 + 28) = 5 × 29 = 145
1)d formula gives the same result: S₁₀ = 10/2 × (2×1 + 9×3) = 5 × (2 + 27) = 5 × 29 = 145.
用 2a + (n − 1)d 公式亦得同样结果：S₁₀ = 10/2 × (2×1 + 9×3) = 5 × (2 + 27) = 5 × 29 = 145。
7. Sum Formula Derivation | 求和公式推导
Understanding why the formula works deepens your mathematical insight. Write the sum of the first n terms forwards and backwards:
理解公式的原理能加深你的数学洞察。将前n项和正写与倒写：
S = 1 + 4 + 7 + ... + (3n − 2)
S = (3n − 2) + (3n − 5) + (3n − 8) + ... + 1
sums to (3n − 1). There are n such columns, so 2S = n × (3n − 1), giving S = n(3n − 1)/2.
将两行垂直相加，每一列的和均为 (3n − 1)。共有 n 列，因此 2S = n × (3n − 1)，即 S = n(3n − 1)/2。
irming our earlier arithmetic. This reasoning underlies the general sum formula and demonstrates a clever pairing strategy.
代入 n = 10：10 × 29 / 2 = 145，验证了之前的计算。这一推理过程正是通用求和公式的本质，并展示了一种巧妙的配对策略。
8. Sum of the First n Terms | 前n项和
= n(3n − 1)/2, we can quickly evaluate sums. For instance, the sum of the first 15 terms:
利用这个数列专属的简化求和公式 Sₙ = n(3n − 1)/2，我们可以快速计算和。例如，前15项的和：
S₁₅ = 15 × (3×15 − 1) / 2 = 15 × (45 − 1) / 2 = 15 × 44 / 2 = 15 × 22 = 330
ifferent arithmetic sequence you must use the general formula or derive a new simplified version.
这个捷径在考试中能节省时间。务必检查求和公式中使用的通项表达式是否与数列匹配。Sₙ = n(3n − 1)/2 这个公式之所以成立，是因为 a = 1 且 d = 3；对于其他等差数列，你必须使用通用公式，或重新推导简化版。
9. Solving Problems Using the Sum | 应用求和解决问题
A typical IGCSE question asks: 'The sum of the first n terms of the sequence 1, 4, 7, ... is 210. Find n.' Set up the equation:
典型的 IGCSE 题目会问："数列 1, 4, 7, ... 的前 n 项和为210，求 n。" 列出方程：
n(3n − 1)/2 = 210 → n(3n − 1) = 420 → 3n² − n − 420 = 0
= 0. Since n must be positive, we take n = 12. Thus, the sum of the first 12 terms is 210.
解这个二次方程。因式分解得 (3n + 35)(n − 12) = 0。由于 n 必须为正数，取 n = 12。因此，前12项的和为210。
Always reject negative or fractional solutions because the number of terms must be a whole positive integer. This blend of sequences and equations is a favourite exam topic.
务必舍弃负数解或分数解，因为项数必须是正整数。这种数列与方程的结合是考试中备受青睐的题型。
10. Arithmetic Means | 等差中项
rithmetic mean is 4, which completes the set 1, 4, 7. This is calculated as (1 + 7)/2 = 4.
两数之间的等差中项是指在构成等差数列时恰好位于中间的那个值。在1与7之间，等差中项为4，从而构成 1, 4, 7。其计算方式为 (1 + 7)/2 = 4。
− 1)/4 = 3, giving the sequence 1, 4, 7, 10, 13. The three inserted means are 4, 7 and 10.
若要在1与13之间插入三个等差中项，实际上就是在构造一个首项为1、末项为13、共计五项的一个等差数列。公差为 (13 − 1)/4 = 3，得到数列 1, 4, 7, 10, 13。插入的三个中项为4、7和10。
11. Real-world Applications | 实际应用
Arithmetic sequences appear in many everyday contexts. Suppose a theatre has 1 seat in the first row, and each subsequent row has 3 more seats than the previous one. The number of seats per row follows our 1, 4, 7, ... pattern.
等差数列出现在许多日常情境中。假设一个剧场第一排有1个座位，之后每一排比前一排多3个座位。每排的座位数就遵循 1, 4, 7, ... 这个模式。
If the theatre has 20 rows, how many seats are there in total? × 59/2 = 590 seats. Problem-solving with sequences makes real-life calculations efficient.
如果该剧场有20排，总共有多少个座位？用求和公式：S₂₀ = 20 × (3×20 − 1)/2 = 20 × 59/2 = 590 个座位。用数列解题能让实际计算高效便捷。
Further applications include savings plans where a fixed amount is added each week, or distances covered in a training programme that increase by a constant gap. Recognising the arithmetic structure unlocks a direct mathematical solution.
更多应用包括每周增加固定金额的储蓄计划，或以恒定差距递增的训练项目所累积的距离。识别出等差数列结构，便可直接运用数学方法求解。
12. Practice and Common Mistakes | 练习与常见错误
Even simple sequences can trap students. Common errors include: confusing n and (n − 1) in the nth term formula; forgetting to halve in the sum formula; using the wrong sign for the common difference; and not checking that n must be an integer when solving term-position problems.
即使简单的数列也可能让学生踩坑。常见错误包括：在第n项公式中混淆 n 与 (n − 1)；在求和公式中忘记除以二；公差的符号使用错误；在求解项的位置问题时未检查 n 必须是整数。
= 3. The sequence is 4, 7, 10, 13, 16, ... - a close relative of our 1, 4, 7, ... family.
练习题：一等差数列的第5项为16，第8项为25。求首项与公差。（提示：根据 a₅ = a + 4d 和 a₈ = a + 7d 列出两个方程。）解得 a = 4，d = 3。该数列为 4, 7, 10, 13, 16, ... - 正是我们 1, 4, 7, ... 家族的近亲。
Master the basic 1, 4, 7, … pattern and you will have a template for tackling a wide range of arithmetic sequence questions throughout the IGCSE Edexcel Mathematics course.
掌握 1, 4, 7, … 这个基本模式，你便拥有了应对 IGCSE Edexcel 数学课程中各类等差数列问题的模板。
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